Asked by Tim
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quote a number.
Answers
Answered by
Damon
LOL 45 degrees
u = Vo cos T
Vi = Vo sin T
at top
v = 0
0 = Vo sin t - g t
t = (Vo/g) sin T where t is time up
time to fall is also t = (Vo/g) sin T
time in air = 2 t = (2 Vo/g) sin T
range = u (2 Vo/g) sin T
range = Vo cos T (2 Vo/g) sin T
Vo and 2 Vo/g are constants given
we need max of sin T cos T
R = k sin T cos T
dR/dT = 0 at max = k[ -sin^2 T + cos^2 T]
or sin T = cos T
that is when sin T = cos T = 1/sqrt2 or T = 45 degrees
u = Vo cos T
Vi = Vo sin T
at top
v = 0
0 = Vo sin t - g t
t = (Vo/g) sin T where t is time up
time to fall is also t = (Vo/g) sin T
time in air = 2 t = (2 Vo/g) sin T
range = u (2 Vo/g) sin T
range = Vo cos T (2 Vo/g) sin T
Vo and 2 Vo/g are constants given
we need max of sin T cos T
R = k sin T cos T
dR/dT = 0 at max = k[ -sin^2 T + cos^2 T]
or sin T = cos T
that is when sin T = cos T = 1/sqrt2 or T = 45 degrees
Answered by
Anonymous
40
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